Question

If $X$ is a binomial variate with mean $\frac{16}{5}$ and variance $\frac{48}{25}$, then $P(X \leq 2) =$

• A. $\frac{3^6 (169)}{5^8}$
• B. $\frac{3^6 (71)}{5^8}$
• C. $\frac{3^8 (43)}{5^8}$
• D. $\frac{3^6 (158)}{5^8}$

Hint:

For binomial distributions, use mean $np$ and variance $npq$ to find $n$ and $p$. Compute $P(X \leq k)$ by summing probabilities $P(X = i)$ for $i = 0$ to $k$.

Answer 1

The Correct Option is A

For a binomial distribution $X \sim \text{Bin}(n, p)$, mean = $np = \frac{16}{5}$, variance = $npq = \frac{48}{25}$. Thus: \[ q = \frac{npq}{np} = \frac{\frac{48}{25}}{\frac{16}{5}} = \frac{48}{25} \cdot \frac{5}{16} = \frac{3}{5} \] \[ p = 1 - q = \frac{2}{5} \] \[ np = n \cdot \frac{2}{5} = \frac{16}{5} \implies n = \frac{16}{5} \cdot \frac{5}{2} = 8 \] So, $X \sim \text{Bin}\left(8, \frac{2}{5}\right)$, $q = \frac{3}{5}$. Compute: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X = k) = \binom{8}{k} \left( \frac{2}{5} \right)^k \left( \frac{3}{5} \right)^{8 - k} \] \[ P(X = 0) = \binom{8}{0} \left( \frac{3}{5} \right)^8 = \frac{3^8}{5^8} \] \[ P(X = 1) = \binom{8}{1} \cdot \frac{2}{5} \cdot \left( \frac{3}{5} \right)^7 = 8 \cdot \frac{2 \cdot 3^7}{5^8} = \frac{16 \cdot 3^7}{5^8} \] \[ P(X = 2) = \binom{8}{2} \cdot \left( \frac{2}{5} \right)^2 \cdot \left( \frac{3}{5} \right)^6 = 28 \cdot \frac{4 \cdot 3^6}{5^8} = \frac{112 \cdot 3^6}{5^8} \] \[ P(X \leq 2) = \frac{3^8 + 16 \cdot 3^7 + 112 \cdot 3^6}{5^8} = \frac{3^6 (3^2 + 16 \cdot 3 + 112)}{5^8} = \frac{3^6 (9 + 48 + 112)}{5^8} = \frac{3^6 \cdot 169}{5^8} \] Option (1) is correct. Options (2), (3), and (4) do not match the computed probability.